# How to solve this equation using $\log$?

I am new to logarithms, and I came across this equation..

$$\ (3)^{4x} - (3)^{(2x + \log_3(12))} +27 = 0 \$$

I need a way to simply things as this seems very complex

Edit:

I tried $$\ (3)^{4x} + 27 = (3)^{(2x + \log_3(12))} \$$

So $$\ (3)^{4x} + 3^{3} = (3)^{(2x + \log_3(12))} \$$

Then, $$\ log ((3)^{4x} + 3^{3}) = (log(3))(2x + \log_3(12)) \$$

• Solve for $y=9^x$. The equation written in terms of $y$ is $y^2-3^{\log_3(12)}y+27=0$. Or $y^2-12y+27=0$. You get that $(y-9)(y-3)=0$. From where either $9^x=y=9$ or $9^x=y=3$. So, either $x=1$ or $x=1/2$. – Olivia Jul 9 '17 at 13:07
• Hint: use the substitution $x=3^{2x}$ to get a quadratic. Also bear in mind: $a^{x+y}=a^xa^y$ and $a^{\log_a(x)}=x$ ($0<x$). – Shuri2060 Jul 9 '17 at 13:07
• The method you're attempting won't get you very far as you've seen as you can't simplify the expression on the LHS easily. – Shuri2060 Jul 9 '17 at 13:09
• Thanks, this is what i was wanting – Ravi Prakash Jul 9 '17 at 13:10
• Is there any formula for $\log(m+n)$ ? – Ravi Prakash Jul 9 '17 at 13:13

$$3^{4x}-3^{2x}\cdot3^{\log_312}+3^3=0$$
$3^{\log_312}$ - is a basic logharithmic rule
$3^{2x} = t$
and solve square equation that depends on $t$